A question that has been bothering me a little while, which Ionescu kindly answered: We know that the Sobolev embedding doesn't hold in the end-point case H^s → L^∞ when s = n/2. The question is, what is the counter-example? Below we give it for the case n=2, s=1, but the method (using Fourier transform) generalizes to arbitrary dimensions.

Suppose f is a function in L², its Fourier transform F is well defined. By Parseval identity, we can write its homogenous H¹ norm as (∫ |ξ|² |F(ξ)|² dξ)^½.

Let F(ξ) = (|ξ|²log(|ξ|))^-1 when |ξ|≥ 2, and let F be constant when |ξ|≤2. F² is easily verified to be integrable in 2 dimensions (it decays faster than |ξ|^-4 near infinity), so its inverse Fourier transform is well defined.

The H¹ norm of f (which is equal to the weighted L² norm of F given above) is also finite. We can check it by writing out the expression in Fourier space in terms of F explicitly, and we see that it boils down to checking the integrability of (|x|log(|x|))^-2 at infinity in 2 dimensions. Writing in polar coordinates, we see that we need to check the integrability of (r (log r)²>)^-1 from 2 to infinity. Its antiderivative is just (-log r)^-1, so the integral is finite.

On the other hand, f is not bounded in L^∞. Since F is a positive function, f(0) is equal to the L¹ norm of F. Again in polar coordinates, we check that the L¹ norm of F depends on the integral from 2 to infinity of the quantity (r log r)^-1. Taking the antiderivative we get (log log r), which diverges at infinity, and hence the integral diverges.